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A Printed Circuit Board (PCB) machine installs integrated circuits onto a board. Before starting to produce a board, the PCB machine requires a 20 minute setup. Once in production, the PCB machine requires only 0.15 minutes per board.

(Round your answer to 2 decimal places.)

a. Currently, the PCB machine produces 500 boards between setups. Given this operating process, what is the capacity of the PCB machine (in boards per min)? ____ boards per minute (Round your percentage answer to 2 decimal places.)

b. Currently, the PCB machine produces 500 boards between setups. Given this operating process, what is the utilization of the PCB machine? _____ % (Do not round intermediate calculations. Round your final answer to 2 decimal places.)

c. Demand for boards occurs at the rate of 2 boards per minute. What is the smallest batch size such that the PCB machine does not constrain the flow rate through the process? _____ batch size

Respuesta :

Answer:

Explanation:

a)

Total time to prosuce 500 boards = 20minute + 0.15*500 = 20+75 = 95minutes

Capacity of the board = Number of boards produced/Time taken to produce the boards = 500/95 = 5.26 boards per minute

b)

Utilization = Production time/(set up time + production time) =

= 75/95 = 0.7894 = 78.94%

c)

2 boards per minute or 1 board per 1/2 minute

Batch size - x

Required cycle time*Batch size = Set up time + production time per board*Bacth size

0.25x = 20 + 0.15x

0.10x = 20

x = 200

Batch size = 200

It should be noted that the capacity of the board will be 5.26 boards per minute.

First, we need to calculate the total time taken which will be:

= 20 minutes + (0.15 × 500)

= 20 + 75

= 95 minutes.

Therefore, the capacity of the board will be;

= 500/95

= 5.26 boards per minute.

The utilization of the machine will be:

= Production time / (Set up time + Production time) × 100

= 75/95 × 100

= 78.94%

Lastly, the smallest batch size will be;

0.25x = 20 + 0.15x

Collect like terms

0.25x - 0.15x = 20

0.10x = 20

x = 20/0.10

x = 200

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